What is the integral from 1 to e of (13+ln(x))^2 ?

The integral from 1 to e of (13+ln(x))^2 is 170e-145

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integral from 1 to e of (13+ln(x))^2

\int_{1}^{e} (13 + ln (x))^{2} d x

170 e - 145

Decimal Notation

317.10791 \dots

Solution steps

\int_{1}^{e} (13 + ln (x))^{2} d x

13 + ln (x) = 13 (1 + \frac{ln ( x )}{13})

= \int_{1}^{e} (13 (1 + \frac{ln ( x )}{13}))^{2} d x

Simplify

(13 (1 + \frac{ln ( x )}{13}))^{2} : 169 + 26 ln (x) + ln^{2} (x)

= \int_{1}^{e} 169 + 26 ln (x) + ln^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{e} 169 d x + \int_{1}^{e} 26 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 169 d x = 169 e - 169

\int_{1}^{e} 26 ln (x) d x = 26

\int_{1}^{e} ln^{2} (x) d x = e - 2

= 169 e - 169 + 26 + e - 2

Simplify

= 170 e - 145

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What is the integral from 1 to e of (13+ln(x))^2 ?
The integral from 1 to e of (13+ln(x))^2 is 170e-145